It is rate for any motion to
1 answer:
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Answer:
0.5A
Explanation:
Using ,
R is the resistance (in Ohms)
V is the voltage (in V)
I is the current (in A)
I = 0.5A
Answer:
a) [volts] = [N m / C],
b) The lines or surface that has the same potential are called equipotential
c) the equipotential lines must also be perpendicular to the electric field lines
Explanation:
a) find the units of the volt
the electric potential energy is
V = k q / r
V = [N m² / C²] C / m
V = [N m / C]
The electric potential is defined as
V = E .s
V = [N / C] [m]
V = [N m / C] = [volt]
we see that in the two expressions the same result is obtained therefore the volt is
[volts] = [N m / C]
b) The lines or surface that has the same potential are called equipotential surfaces, the great utility of these lines or surfaces is that a face can be displaced on it without doing work.
c) The electric potential is defined as the gradient of the electric field
v =
therefore the equipotential lines must also be perpendicular to the electric field lines
Answer:
a) = 0.25 m / s b) u = 0.25 m / s
Explanation:
a) To solve this problem let's start with the conservation of the moment, for this we define a system formed by the ball plus the dog, in this case all the forces are internal and the moment is conserved
We will write the data
m₁ = 0.40 kg
v₁₀ = 9.0 m / s
m₂ = 14 kg
v₂₀ = 0
Initial
po = m₁ v₁₀
Final
= (m₁ + m₂) vf
po = pf
m₁ v₁₀ = (m₁ + m₂)
= v₁₀ m₁ / (m₁ + m₂)
= 9.0 (0.40 / (0.40 +14)
= 0.25 m / s
b) This is the reference frame of the center of mass of the system in this case the speed of this frame is the speed of the center of mass
u = 0.25 m / s
In the direction of movement of the ball
c) Let's calculate the kinetic energy in both moments
Initial
K₀ = ½ m₁ v₁₀² +0
K₀ = ½ 0.40 9 2
K₀ = 16.2 J
Final
= ½ (m₁ + m₂)
2
= ½ (0.4 +14) 0.25 2
= 0.45 J
ΔK = K₀ -
ΔK = 16.2-0.445
ΔK = 1575 J
These will transform internal system energy
d) In order to find the kinetic energy, we must first find the velocities of the individual in this reference system.
v₁₀’= v₁₀ -u
v₁₀’= 9 -.025
v₁₀‘= 8.75 m / s
v₂₀ ‘= v₂₀ -u
v₂₀‘= - 0.25 m / s
‘=
- u
= 0
Initial
K₀ = ½ m₁ v₁₀‘² + ½ m₂ v₂₀‘²
Ko = ½ 0.4 8.75² + ½ 14.0 0.25²
Ko = 15.31 + 0.4375
K o = 15.75 J
Final
= ½ (m₁ + m₂) vf’²
= 0
All initial kinetic energy is transformed into internal energy in this reference system